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Mirrors > Home > MPE Home > Th. List > ddif | Structured version Visualization version GIF version |
Description: Double complement under universal class. Exercise 4.10(s) of [Mendelson] p. 231. (Contributed by NM, 8-Jan-2002.) |
Ref | Expression |
---|---|
ddif | ⊢ (V ∖ (V ∖ 𝐴)) = 𝐴 |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | vex 3477 | . . . . 5 ⊢ 𝑥 ∈ V | |
2 | eldif 3958 | . . . . 5 ⊢ (𝑥 ∈ (V ∖ 𝐴) ↔ (𝑥 ∈ V ∧ ¬ 𝑥 ∈ 𝐴)) | |
3 | 1, 2 | mpbiran 706 | . . . 4 ⊢ (𝑥 ∈ (V ∖ 𝐴) ↔ ¬ 𝑥 ∈ 𝐴) |
4 | 3 | con2bii 357 | . . 3 ⊢ (𝑥 ∈ 𝐴 ↔ ¬ 𝑥 ∈ (V ∖ 𝐴)) |
5 | 1 | biantrur 530 | . . 3 ⊢ (¬ 𝑥 ∈ (V ∖ 𝐴) ↔ (𝑥 ∈ V ∧ ¬ 𝑥 ∈ (V ∖ 𝐴))) |
6 | 4, 5 | bitr2i 276 | . 2 ⊢ ((𝑥 ∈ V ∧ ¬ 𝑥 ∈ (V ∖ 𝐴)) ↔ 𝑥 ∈ 𝐴) |
7 | 6 | difeqri 4124 | 1 ⊢ (V ∖ (V ∖ 𝐴)) = 𝐴 |
Colors of variables: wff setvar class |
Syntax hints: ¬ wn 3 ∧ wa 395 = wceq 1540 ∈ wcel 2105 Vcvv 3473 ∖ cdif 3945 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1796 ax-4 1810 ax-5 1912 ax-6 1970 ax-7 2010 ax-8 2107 ax-9 2115 ax-ext 2702 |
This theorem depends on definitions: df-bi 206 df-an 396 df-tru 1543 df-ex 1781 df-sb 2067 df-clab 2709 df-cleq 2723 df-clel 2809 df-v 3475 df-dif 3951 |
This theorem is referenced by: complss 4146 dfun3 4265 dfin3 4266 invdif 4268 ssindif0 4463 difdifdir 4491 |
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